// 14、正确定义axjx 8:51
function Ajax(url:string,callback:any,type:string='get'){
           const xhr=new XMLHttpRequest()
           xhr.open(type,url,true)
           xhr.send()
           xhr.onreadystatechange=()=>{
               if(xhr.status===200&&xhr.readyState===4){
                   return callback(JSON.parse(xhr.response))
               }
           }
}